Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $r = \dfrac{-2t - 2}{t + 6} \div \dfrac{t^2 - t - 2}{-3t^2 - 33t - 90} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{-2t - 2}{t + 6} \times \dfrac{-3t^2 - 33t - 90}{t^2 - t - 2} $ First factor out any common factors. $r = \dfrac{-2(t + 1)}{t + 6} \times \dfrac{-3(t^2 + 11t + 30)}{t^2 - t - 2} $ Then factor the quadratic expressions. $r = \dfrac {-2(t + 1)} {t + 6} \times \dfrac {-3(t + 6)(t + 5)} {(t + 1)(t - 2)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {-2(t + 1) \times -3(t + 6)(t + 5) } {(t + 6) \times (t + 1)(t - 2) } $ $r = \dfrac {6(t + 6)(t + 5)(t + 1)} {(t + 1)(t - 2)(t + 6)} $ Notice that $(t + 1)$ and $(t + 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {6(t + 6)(t + 5)\cancel{(t + 1)}} {\cancel{(t + 1)}(t - 2)(t + 6)} $ We are dividing by $t + 1$ , so $t + 1 \neq 0$ Therefore, $t \neq -1$ $r = \dfrac {6\cancel{(t + 6)}(t + 5)\cancel{(t + 1)}} {\cancel{(t + 1)}(t - 2)\cancel{(t + 6)}} $ We are dividing by $t + 6$ , so $t + 6 \neq 0$ Therefore, $t \neq -6$ $r = \dfrac {6(t + 5)} {t - 2} $ $ r = \dfrac{6(t + 5)}{t - 2}; t \neq -1; t \neq -6 $